The ionic radius of Na+ ions is 1.02 Å. The ionic radii (in Å) of Mg2+ and Al3+, respectively, are:

🧠 All three ions are isoelectronic with 10 electrons — radius shrinks as Z rises $\ce{Na^+}$ ($Z=11$), $\ce{Mg^{2+}}$ ($Z=12$), $\ce{Al^{3+}}$ ($Z=13$). Same 10-electron cloud, just more protons pulling inward. Each extra proton tightens the cloud noticeably.

🗺️ Apply the Z effect $\ce{Na^+}$ ($Z=11$): given as $1.02\,\text{Å}$. $\ce{Mg^{2+}}$ ($Z=12$): one more proton than $\ce{Na^+}$ — must be smaller. Standard tabulated value: $0.72\,\text{Å}$. $\ce{Al^{3+}}$ ($Z=13$): two more protons than $\ce{Na^+}$ — must be even smaller. Standard tabulated value: $0.54\,\text{Å}$.

Now match the options: (a) 1.05, 0.99 — both are LARGER than $\ce{Na^+}$. Impossible for isoelectronic ions with higher Z. ✗ (b) 0.72, 0.54 — both smaller, decrease as Z rises. ✓ (c) 0.85, 0.99 — Al³⁺ given as larger than Mg²⁺. Backwards. ✗ (d) 0.68, 0.72 — Al³⁺ shown larger than Mg²⁺. Wrong direction. ✗

⚠️ The trap Options (c) and (d) violate the Z rule outright (Al³⁺ shown bigger than Mg²⁺), but students sometimes mark them by remembering "more positive = different periods" without realising all three are in the same isoelectronic set. In an isoelectronic group, the ion with the most positive charge is always the smallest. Always check that the size order matches the Z order before locking in the answer.

$\boxed{\text{Answer: (b)}}$